Two Plus Two Poker Forums Probability Interview Question with an Asset Manager
 Register FAQ Search Today's Posts Mark Forums Read Video Directory TwoPlusTwo.com

 Notices 2013 Two Plus Two Party at South Point BET RAISE FOLD screening at 3PM in lounge next to poker room. July 6th Party registration at 5PM. Food served at 6PM

 Probability Discussions of probability theory

 07-22-2012, 08:17 AM #16 Carpal \'Tunnel     Join Date: Mar 2010 Location: Philadelphia Posts: 6,128 Re: Probability Interview Question with an Asset Manager I don't think that's what the question meant. It was more like an all or nothing scenario, but what is the better choice as far as probabilities go. This is the type of question I'd be somewhat scared of in an interview. It's not all that difficult with time, but with someone staring at you or you are on the spot, it tends to be a lot harder. I've had simple questions asked of me and sometimes I freeze up for a second with 5 thoughts in my head at the same time "What does he really mean here? Is this a trick question? What answer will impress him the most? How'd she look naked?" etc. I usually get through it, but I'd be lying if I said I never bombed an interview question.
 07-22-2012, 09:15 AM #17 Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,450 Re: Probability Interview Question with an Asset Manager The confusing part to me was that I didn't recognize that if you wait until the end you EARN \$10, whereas if you bet before your expectation is to EARN can only be a maximum of \$9 (b/c you have to pay the \$1 to play). When I first read the problem it sounded to me like the maximum possible was only \$9 earned regardless of the choice.
07-22-2012, 09:17 AM   #18
Carpal \'Tunnel

Join Date: Jun 2005
Location: Psychology Department
Posts: 7,450
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by Didace Don't you get to keep betting if you win? This seems to imply that you do.
And this was the other thing I thought was possible when I read the problem. I assumed if you won a bet, you could keep playing until all of the marbles were gone, with the last marble always being a freeroll. If this was the case, there would clearly be a better strategy if the marbles fell in your favor at some point (if they never fall in your favor, then just wait until the end).

07-22-2012, 11:40 AM   #19
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,953
Re: Probability Interview Question with an Asset Manager

Quote:
Originally Posted by Didace
Don't you get to keep betting if you win? This seems to imply that you do.
Quote:
 Originally Posted by mburke At any point you can bet 1\$, if it hits red you win you win 9\$ plus your 1\$ back. If it hits blue you lose and are no longer allowed to wager because you're broke.
The very thing you quoted implies that you only get to bet once. Otherwise, you wouldn't necessarily be broke.

07-22-2012, 11:43 AM   #20
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,953
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by Sherman The confusing part to me was that I didn't recognize that if you wait until the end you EARN \$10, whereas if you bet before your expectation is to EARN can only be a maximum of \$9 (b/c you have to pay the \$1 to play). When I first read the problem it sounded to me like the maximum possible was only \$9 earned regardless of the choice.

"However, you can also choose to not bet and not pay your 1\$. If you do not bet, and the final marble drawn is red, you win 10\$, if it is blue, you win 0\$."

07-22-2012, 01:38 PM   #21
veteran

Join Date: Nov 2009
Posts: 2,770
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by BruceZ The very thing you quoted implies that you only get to bet once. Otherwise, you wouldn't necessarily be broke.
I have \$1. If I bet and lose I now have \$0 and can no longer bet because I am broke. If I bet and win I now have \$10 and am not broke. Why can't I bet with that money? This is how investing would work in reality.

07-22-2012, 01:49 PM   #22
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,953
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by Didace I have \$1. If I bet and lose I now have \$0 and can no longer bet because I am broke. If I bet and win I now have \$10 and am not broke. Why can't I bet with that money? This is how investing would work in reality.
That's not how the marble futures market works. You can only make 1 investment per 100 draws. This is due to new SEC regulations.

07-22-2012, 02:22 PM   #23
Carpal \'Tunnel

Join Date: Mar 2008
Location: Ruling with an iron pocketbook.
Posts: 10,663
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by wil318466 Umm.. Just a guess, but if you win 9 dollars if you bet, but win 10 dollars if you get to the final marble, my guess would be that you have to have odds overcome the difference between the profit of 9 dollars vs 10 dollars, so your odds would have to be a better expected value to win 9 dollars than 50/50 to win 10. Or am I completely wrong?
this is a correct way of thinking about it. try and think of it in terms of expected profit (which is a function of expected revenue - expected cost). the part after is slightly incorrect. hint: it involves the one dollar.

Quote:
 Originally Posted by randomwalk I'm not sure I'm interpreting the question correctly, but I think the answer is that it doesn't matter what you do. Your expected earnings is \$5. This is because every strategy is equivalent to picking the last marble, which has a 50% chance of being red a priori.
this is incorrect. but i agree the question is very suspectly worded. this was how it was read to me verbatim however so i figured it might be fun for you to try and solve it. i will answer any questions i can just as he did answer my questions.

this is not the answer though.

think about extreme (but simple) cases, where for example there were 2 red marbles, 1 blue marble left.

Quote:
 Originally Posted by wil318466 But what if 10 red and 10 blues are taken out, (down to 80 marbles), then 5 blues get taken out in a row? The percentages have now changed, and when you pick you only get paid 9, but if you don't play, you have a chance to win 10. Like you said, I'm not sure if I'm interpreting this correctly (the post is a bit hard to follow in all honesty)
you are correct that when you place a bet you can only win 9 plus your 1 back. aside from that i'm not sure i understand the question.

the answer is there are scenarios in which all blues are taken, or all reds are taken. but these can be accounted for with probability distributions. good work, this is how i envisioned it initially (what happens if 10 and 10 are taken?)

think of it as the sample size shrinks even further and your edge approaches would would be an "ideal" scenario.

in these sorts of logic problems i have noticed that its usually best to think about the extreme cases one way or another as the solutions.

07-22-2012, 02:35 PM   #24
Carpal \'Tunnel

Join Date: Mar 2008
Location: Ruling with an iron pocketbook.
Posts: 10,663
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by BruceZ Nice problem. The best strategy is to never bet. In that case your EV is \$5. All other strategies have an EV less than \$5.
this is the incorrect value, but i like your line of reasoning. expound on it!

Quote:
 You might think you can do better if you bet when the percentage of red marbles is greater than 60% since when that condition arises your EV would be greater than \$5. However, your EV would be even better by not betting because the last marble has the same probability of being red as the next marble, but you get paid more by not betting. As an extreme example, suppose all of the remaining marbles were red. You still shouldn't bet because you will only make \$9 instead of \$10 by waiting until the end.
this is in essence the solution to half of the problem which i found (along with the EV of bet/no-bet). the other half is calculating the expected profit from either solution (bet or no bet).

he wouldn't tell me what the extension of the solution was but i can tell you that the EV of profit in the scenario of not betting i got (in for example, 2red, 1 blue remaining) was \$6.67(no bet) and \$5.67 (bet). to do this i just made a probability distribution tree and calculated the possible scenarios:

bet no bet
.66(r)*9, .33(b)*-1 .66(r) [.5red, .5blue] .33(b) [red 1.0]

my theory is basically in line with bruce, in that the nearest you can approach the guaranteed 1\$ would only equal it. (for example if all blues were taken).

even in this extreme scenario you would have an expected profit of (if betting) 9\$, vs not betting (10\$).

in addition, due to the random sampling, i theorize that it would hypothetically never be optimal to bet vs wait because you would expect if you ran this game millions of times for the distribution of the drawn marbles to approach .5r/.5b. in that scenario it is undoubtedly profitable to not bet.

in addition: to those who asked about whether you can bet once. bruce is correct, you can only bet once. i struggled with this (thinking it would be an investment, like soebody had said, where we could maximize a shot once and then continue to bet if that edge remained. however, it is as bruce called it regulated such that you can ONLY BET ONCE. this was again something i had to ask.

there was another qeustion fromt he interview concerning options that was fun that i might post here (right before this question)

thanks guys! let me know if you have any other different answers or see holes in my logic.

07-22-2012, 02:51 PM   #25
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,953
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by mburke05 this is the incorrect value, but i like your line of reasoning. expound on it!
It's not incorrect. The EV for the game when you never bet is \$5. That's trivial because the last marble has probability 1/2 of being red.

There will be times after some marbles are drawn that your EV for waiting will be greater than \$5. It can be as much as \$10 when only reds are left. But the EV of waiting will always be greater than the EV of betting because whatever the probability is that the next marble is red, the probability that the last marble is red is the same, and we don't have to bet \$1 on the last marble. That's the key point, that the probability for the last marble and the next marble are the same. There are no calculations necessary to reach this conclusion. I've already given the complete argument.

Quote:
 this is in essence the solution to half of the problem which i found (along with the EV of bet/no-bet).
No, the 60% solution is completely irrelevant.

Quote:
 he wouldn't tell me what the extension of the solution was but i can tell you that the EV of profit in the scenario of not betting i got (in for example, 2red, 1 blue remaining) was \$6.67(no bet) and \$5.67 (bet). to do this i just made a probability distribution tree and calculated the possible scenarios:
Of course it's \$6.67. That's because the probability that the last one is red is 2/3. In the original game, it's 1/2, so the EV is \$5. You don't need a decision tree. You're missing the key.

Last edited by BruceZ; 07-22-2012 at 03:17 PM.

 07-22-2012, 03:27 PM #26 Carpal \'Tunnel     Join Date: Mar 2008 Location: Ruling with an iron pocketbook. Posts: 10,663 Re: Probability Interview Question with an Asset Manager Bruce, part of the solution was calculating the expected profit in these various strategies, hence the calculations.
07-22-2012, 03:35 PM   #27
centurion

Join Date: Feb 2010
Posts: 104
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by mburke05 Bruce, part of the solution was calculating the expected profit in these various strategies, hence the calculations.
Bruce is trying to tell you that the calculation of the EV is extremely simple, only depending on the probability that the last marble is red.

 07-22-2012, 04:00 PM #28 Carpal \'Tunnel     Join Date: Mar 2008 Location: Ruling with an iron pocketbook. Posts: 10,663 Re: Probability Interview Question with an Asset Manager gotcha kk.
07-22-2012, 08:15 PM   #29

Join Date: Jan 2006
Posts: 714
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by BruceZ The very thing you quoted implies that you only get to bet once. Otherwise, you wouldn't necessarily be broke.
Quote:
 At any point you can bet 1\$, if it hits red you win you win 9\$ plus your 1\$ back. If it hits blue you lose and are no longer allowed to wager because you're broke.
Why should losing necessarily mean you lose your chance at the \$10 at the end. That chance does not require you to place a wager. Nowhere is it explicitly stated that you cannot continue. It only says you can no longer wager.

In more than one way, this problem is very poorly stated. A lot of people thought you could bet more than once including the OP at first.

07-22-2012, 08:36 PM   #30
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,953
Re: Probability Interview Question with an Asset Manager

Quote:
 Originally Posted by R Gibert Why should losing necessarily mean you lose your chance at the \$10 at the end. That chance does not require you to place a wager. Nowhere is it explicitly stated that you cannot continue.
Quote:
 Originally Posted by mburke05 However, you can also choose to not bet and not pay your 1\$. If you do not bet, and the final marble drawn is red, you win 10\$, if it is blue, you win 0\$.
I don't know what you people's problem is. Seems crystal clear to me.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules

All times are GMT -4. The time now is 06:28 AM.

 Contact Us - Two Plus Two Publishing LLC - Privacy Statement - Top